Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How many numbers n between 1 and 600 do exist such that the floor of $\sqrt[4]{n}$ divides n. please solved the problem in a general case.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Note that $\lfloor \sqrt[4]n \rfloor=2$ for $16 \le n \le 80$ so how many $n$'s in this range qualify?

If we let $k=\lfloor \sqrt[4] n \rfloor$ we have $\sum_{i=1}^{k-1}\left(\lfloor \frac {(i+1)^4-(i^4+2)}i\rfloor+1\right)+\lfloor \frac {n-(k^4+1)}k\rfloor+1$ for the general case. The idea is as the example above, to count the number of multiples of $i$ from $i^4$ through $(i+1)^4-1$. Note that we start and end with a multiple of $i$.

share|improve this answer
    
There is 23 n that qualify in this range. –  Emad Feb 4 at 22:59
    
I updated the range-it should go through $80$, so $23$ is not correct. My fault. Since $\lfloor \sqrt[4]{600}\rfloor=4$ you only have three more ranges to worry about. –  Ross Millikan Feb 4 at 23:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.